Shell Programming and Scripting

Taking inputs in the script which is space separated and passing this to a function and i have assigned like below
And then when I use for loop for the inputs i got from the user, it is taking only the first argument.

Code:

Enter Names : Bala Sundar Sridhar
read names
namesCheck $names

function namesCheck() {
myname=$1

for i in myname
do
printf "\n$i"
done
}

this is printing always the first name .. here as Bala

Pl help me how to get all values in the function.

Hi,
If you want to print all the arguments the use $@

function namesCheck()
{
echo $@
}

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You can fix your code by putting double quotes around the variable names when calling your function

Code:

read names
namesCheck "$names"

function namesCheck() {
myname=$1

for i in $myname
do
printf "\n$i"
done
}

use for loop to call function namesCheck, not in function namesCheck to use for loop.

Code:

function namesCheck
{
  printf "\n$1"
}

for name in `echo "Bala Sundar Sridhar"`
do
  namesCheck $name
done

How about getopts (see lower part of the page linked) ...?

The reason is that <space> is a special character to the shell: it is the "field separator". If you feed a command (and a shell function in this regard is similar to a command) some arguments they get plucked from the commandline one by one and the shell supposes them to be separated by blanks:

Code:

command arg1 arg2 arg3

"command" will suppose "arg1" to be the first argument, "arg2" to be the second, and so on. If it takes only 2 arguments it will simply ignore the rest . this is what has happened to your other names.

What to do if you want to pass an argument which contains space characters? This is where quoting comes in.

Code:

command "arg1 arg2" arg3

Now "command" will suppose "arg1 arg2" to be the first argument and "arg3" to be the second.

No let us examine your script: with the line

Code:

namesCheck $names

you feed the function namesCheck 3 arguments, which it gladly receives. The first line in namesCheck() is

Code:

myname=$1

which means: take the first argument and store it in $myname. Exactly this is what the shell does - it stores the first argument passed - the first name - into $myname.

To solve this problem you have two possibilities: either pass all the names as one argument, then you could use the for-loop like you did. For this you will have to surround $names by double quotes when you call namesCheck():

Code:

nameCheck "$names"

Or you could pass the names as separate arguments, but then the function has to have provisions to deal with multiple arguments, like pravin27 mentioned.

I hope this helps.

bakunin